I have a table called "reports"
it looks like:
user_id | report_post
1 2
2 2
3 2
4 10
Now I want to list the first three entries at first, cause the reported post id "2" is 3 times in this table... i want so sort them by the maximum of entries.
Hope you guys understand... thanks alot
----edit------
my output have to look like this
report_post | entries
2 3
10 1
Select report_post, Count(1) As Entries
From reports
Group By Report_Post
Order By Count(1) DESC
SELECT *
FROM (
SELECT user_id, report_post, COUNT(*) AS cnt
FROM reports
GROUP BY report_post
) c
ORDER BY cnt DESC
With your edit, this does what you're asking:
select report_post, count(*) entries
from reports
group by report_post
order by entries desc
Use a subquery. This should do the trick in MySQL:
select * from reports
order by (
select count(*) from reports as reports_a
where reports_a.report_post=reports.report_post
) desc;
(the above answer's your question before you edited it to change it's meaning)
For the edited question, it is a trivial example of a group by:
select report_post,count(*) as entries
from reports
group by report_post
order by entries desc;
SELECT report_post as report_post, count(report_post) as entries
FROM `reports` group by `report_post`
Single query.
SELECT * FROM reports ORDER BY entries DESC
I'm a little unsure of what exactly you are asking. Are you just trying to get it to return entries where the report_post id is equal to "2"?
If that is the case, this should work:
SELECT * FROM reports WHERE report_post=2;
Sorry if I misunderstood your question.
Related
I have a table named items which has 3 columns : id, user_id, item_name.
I want to select and show all users that have most submitted items in that table.
For instance :
User-1 has 3 items,
User-2 has 8 items,
User-3 has 5 items,
User-4 has 8 items, and
User-5 has 8 items too.
Based on what I need, the query should be outputting User-2, User-4 and User-5.
My knowledge of MySQL is not thorough unfortunately and I can't get this done by myself.
Your help is much appreciated.
EDIT #1 :
Here's the query that I tried and didn't output my desired result :
SELECT COUNT(id) AS total_count
, user_id
FROM ".DB_PREFIX."items
GROUP
BY user_id
It shows all users and their total number of items submitted. As I mentioned earlier, I need all top users.
E.g.:
SELECT a.*
FROM
( SELECT user_id
, COUNT(*) total
FROM my_table
GROUP
BY user_id
) a
JOIN
( SELECT COUNT(*) total
FROM my_table
GROUP
BY user_id
ORDER
BY total DESC
LIMIT 1
) b
ON b.total = a.total;
I'm trying to figure out how to do this on my own but it looks like a dead end to me.
I am working under wordpress framework and with some custom tables.
The result i am trying to achieve is very simple but the way to get there is just too much for my head right now.
I need to select the top 50 results from tableOne based on the ammount of times that the id from tableOne is mentioned in tableTwo under some simple where conditions.
Using $wpdb class for the latest WordPress build, what can i use to achieve this?
Thanks
This is the Simple tableOne Query to get all posts:
$allPosts = $wpdb->get_results("SELECT * FROM pokeGrid_images WHERE status='0' ORDER BY tempo DESC LIMIT ".$limit." OFFSET ".$offset."");
Now i need the first 50 results from this table, based on the number of times their id is mentioned on the second table with this structure: http://prntscr.com/byz2qw
Edit:
http://prntscr.com/byz79b
Note: Basically this is a forum, table one has the posts, table 2 the upvotes.
The expression must gather the most upvoted posts for the last X days.
If it was 1 post the expression would be Select Count(*) FROM tableTwo Where tempo > then ".$variableWithUnixTimeDIff."
column tempo is a now() timestamp.
Thanks
Note: Basically this is a forum, table one has the posts, table 2 the
upvotes. The expression must gather the most upvoted posts for the
last X days. If it was 1 post the expression would be Select Count(*)
FROM tableTwo Where tempo > then ".$variableWithUnixTimeDIff."
SELECT *, Sum(score) AS totalScore FROM tableTwo INNER JOIN tableOne ON tableTwo.memeID = tableOne.id GROUP BY memeID ORDER BY totalScore DESC;
Try this query.
//Edit - I just created a sample table with my own data. I only created the tableTwo, where this query did work:
SELECT *, Sum(score) AS totalScore FROM tableTwo GROUP BY memeID ORDER BY totalScore DESC
So after that, just inner join data from tableOne and it will work!
I'm trying to select the latest 10 records from multiple tables (ORDER BY date). For example, 8 of the newest records might be in one table and 2 in another (10 rows in total). Is there a way to select those 10 records?
SELECT *
FROM
( SELECT * FROM x
UNION ALL
SELECT * FROM y
) n
ORDER
BY date DESC
LIMIT 10;
You can maybe use:
SELECT column_name(s)
FROM table1
ORDER BY date LIMIT 0,8
UNION ALL
SELECT column_name(s)
FROM table2
ORDER BY date LIMIT 0,2;
SELECT * FROM (
SELECT some_data AS alias1, date_field AS mydate
FROM table1
UNION ALL
SELECT datazzz AS alias1, another_datefield AS mydate
FROM table2
)
ORDER BY mydate DESC LIMIT 10
Syntax might need a little bit of tweaking, but that's the gist of it.
Specifically, you need to select whatever data you want out of each of the tables and then use aliases to make sure they have the same column names (otherwise they can't be returned in the same result set). Then after that you need to order by the common date field.
I have had a look through google results, but can't find a simple answer, I have a table
id, points (and more fields, but they dont affect anything)
How would I go about Getting position of record 24(id) ordered by points DESC?
I'm not sure I've understood your question
select * from table order by points desc limit 23,1
Select Count(*) from (
Select ID from UnNamedTable A where A.points>(Select B.Points from UnNamedTable B where B.ID=24)
)
Are you trying to figure out out what place a user is in based on a points system?
If so then just find how many users have more points than a particular user then add 1.
SELECT COUNT(*) + 1 AS 'position' FROM table WHERE points >
( SELECT points FROM table WHERE id = someIDhere )
You could create a variable to keep track of your row position after sorting. Something like this would work:
select #rownum:=#rownum+1 ‘rank’, p.* from table1 p, (SELECT #rownum:=0) ID order by Points desc limit 10;
Welcome,
I'm wondering is it possible to reverse returned data in sorting "order by desc" but i want that data in reverse order.
For example, i got table with values
ID
1
2
3
4
And i do
Order by ID ASC LIMIT 3
I got
1
2
3
When i do Order by ID DESC limit 3
i get
4
3
2
I would like to have
3
2
1
So i would like to order by ASC but revers results.
I was always doing this in PHP side using array_reverse, but today i want ask You.
Maybye i'm wrong and i can do this just in Mysql.
Regards
SELECT *
FROM (
SELECT ...
FROM ...
ORDER BY ID ASC
LIMIT 3
) AS sq
ORDER BY ID DESC
Think of it as working in two steps. First it executes the inner query: selects 3 records with lowest IDs. Then in the outer query it sorts them in descending order.
You can fetch the first three rows using a subquery and then reverse the order of these rows in an outer query:
SELECT *
FROM
(
SELECT *
FROM yourtable
ORDER BY ID
LIMIT 3
) T1
ORDER BY ID DESC
Nope, you aren't wrong. There are no difference for any sensible amount of data. Array_reverse is all right.
That's not a thing you have to be concerned too much of. Just use whatever you like more - for readability or other subjective reasons
You can do this with a sub query :
SELECT * FROM
(SELECT * FROM myTable ORDER BY idMyTable LIMIT 0, 3) AS r
ORDER BY r.idMyTable DESC
Resources :
Mysql - Subqueries
You could use an outer SELECT to reverse the order:
SELECT *
FROM (
SELECT …
ORDER BY id ASC
LIMIT 3
) sub
ORDER BY id DESC